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The final exam for a chemistry 201 course, consisting of multiple-choice questions and redox reaction balancing problems. Students are required to fill in the blanks with the correct answers and balance the given redox reactions. Topics covered include atomic orbitals, compounds' names, structures, and properties, as well as redox reactions and their balancing.
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Chem 201 Name (PRINT!) KEY
Final Exam December 9, 2008
Hybrid atomic orbitals named sp2 are used by the central atom when it has 3
regions of electron density in the Lewis dot diagram.
An aldehyde would react with dichromate ion to yield the organic functional group called
Carboxylic acid (Be as specific as possible.)
A ketone would react with sodium borohydride to yield the organic functional group
Secondary alcohol (Be as specific as possible.)
63
2 + has 29 protons, 27 electrons, and 34 neutrons.
T, V, E, and H are examples of state functions, while q and w are
examples of path functions.
The quantum number n is called the primary quantum number; one of the
names of l is the secondary/azimuthal/subsidiary quantum number.
Oxides of metalloids are most likely to dissolve in solution described as
Acidic or basic (acidic/basic/either acidic or basic).
The energy released when a neutral gas phase atom gains an electron is called its
Electron affinity
Of the salts LiF, NaCl. and KBr, the one having the largest value for lattice energy would
be LiF.
Sr 3 (PO 3 ) 2 strontium phosphite
W(ClO 3 ) 3 tumgsten(III)chlorate
Hg 2 (C 2 H 3 O 2 ) 2 mercurous acetate
H 3 AsO 3 (aq) arsenous acid
Chem 201 A Final Exam, p 2 Name
O
OH
pentanoic acid
HO
2-butanol
NH 2
O
propanamide
methyl butanoate
O
O
cyclohexanone
O
Joules has
a frequency of:
h
2.50x
6.626x
= 3.77 x 10
16 s
a wavelength of:
λ =
c
ν
2.9979x
8 m s
3.77 x
16 s
= 7.95 x
Chem 201 A Final Exam, p 4 Name
acceptable Lewis dot diagram, using o or x to represent every valence electron. If
resonance occurs, only one of the forms need be shown. Then indicate the electronic
geometry around the central atom, and the overall molecular geometry.
Hybrid orbitals for S = sp
Electronic geometry = tetrahedral
Molecular geometry = bent
TeF 5
Hybrid orbitals for Te = sp3d
Electronic geometry = octahedral
Molecular geometry = square pyramidal
N = A = S =
Hybrid orbitals for I = sp3d
Electronic geometry = trigonal bipyramidal
Molecular geometry = see-saw
Chem 201 A Final Exam, p 5 Name
molar enthalpy change is termed
ΔHf
o )
4 C(s) + 5 H 2 (g) = C 4 H 10 (g)
(b) Write the balanced combustion reaction of one mole of butane gas with oxygen gas to
form carbon dioxide gas and liquid water.
C 4 H 10 (g) + 6.5 O 2 (g) = 4 CO 2 (g) + 5 H 2 O(l)
The molar enthalpy change of combustion of butane
o is -2878 kJ/mole.
The
ΔHf
o of carbon dioxide gas is -393.5 kJ/mole, and the
ΔHf
o of liquid water
is -285.83 kJ/mole. Calculate the value of
o for butane gas.
ΔHreaction = Sum ΔH products – Sum of ΔH reactants
-2878 kJ/mole = [4(-393.5 kJ/mol) + 5(-285.83 kJ/mol)] – [ΔHf butane + 6.5 (0)]
ΔHf
o butane gas = +2878 - 3003 = - 125 kJ/mole
2 NaOH(s) + Cu
2+ (aq) = 2 Na
(aq) + Cu(OH) 2 (s)
Chem 201 A Final Exam, p 7 Name
results in the form of a table is strongly encouraged. Assuming complete reaction, how
many moles remain at the end of the reaction for each reactant and each product when
40.00 g of phosphorous, 75.00 g of barium hydroxide, 20.00 g of water, and 15.00 g of
sulfuric acid are reacted according to the reaction:
2 P 4 + 3 Ba(OH) 2 + 6 H 2 O + H 2 SO 4
3 BaSO 4 + 6 H 3 PO 2 + 2 PH 3
40.00 g P4(1 mole/123.896g) = 0.3228 mol P
75.00 g Ba(OH)2 (1 mole/171.3416 g) = 0.4377 mole Ba(OH)
20.00 g H2O(1 mole H2O/18.01528 g) = 1.110 mole H2O
15.00 g H2SO4 (1 mol H2SO4/98.0794 g) = 0.1529 mole H2SO
2 P 4 + 3 Ba(OH) 2 + 6 H 2 O + H 2 SO 4
__
3 BaSO 4 + 6 H 3 PO 2 + 2 PH 3
mol inti .3228 .4377 1.110 .1529 0 0 0
mol react 2x 3x 6x x - - -
mol form - - - - 3x 6x 2x
left .3228- .4377- 1.110- .1529-
2x=0 3x=0 6x = 0 x = 0
x=? .1614 .1459 .1850.
Left .0310 0 .235 .0070 .4377 .8754.
Chem 201 A Final Exam, p 8 Name
photon associated with this transition is absorbed (absorbed/emitted).
Calculate the wavelength, in nm, of this photon.
€
1
λ
= R H
1
n f
2
1
n i
2
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
= 1.09737m
-1 1
5
2
1
3
2
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ = 7.803 x 10
5 m
€
λ =
1
7.803x
5 m
10
9 nm
1 m
⎛
⎝
⎜
⎞
⎠
⎟ =^ 1281 nm
heat 4.18 J/g K) the cup contained were both initially at a temperature of 21.
o C.
A 50.0 g piece of lead, initially at a temperature of 175.
o C, was added to the water and
cup. The final common temperature was 28.
o C. What is the specific heat of lead?
0 = (15.0 J/K)(28.8-21.0deg) + (25.0g)(4.28J/gK)(28.8-21.0deg)
0 = 117 J + 815 J - 7310 g K (SpHt)
SpHt = 0.127 J/gK